AP Chemistry Chapter 17 Additional Aspects of Aqueous Equilibria
Chapter 17. Additional Aspects of Equilibrium
Sample Exercise 17.1 (p. 720)
What is the pH of a solution made by adding 0.30 mol of acetic acid (HC2H3O2) and 0.30 mol of sodium acetate (NaC2H3O2) to enough water to make 1.0 L of solution?
(4.74)
Practice Exercise 17.1
Calculate the pH of a solution containing 0.085 M nitrous acid (HNO2, Ka = 4.5 x 10-4) and 0.10 M potassium nitrite (KNO2).
(3.42)
Sample Exercise 17.2 (p. 722)
Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.
([F-] = 1.4 x 10-3 M; pH = 1.00)
Practice Exercise 17.2
Calculate the formate ion concentration and pH of a solution that is 0.050 M in formic acid (HCOOH;
Ka = 1.8 x 10-4) and 0.10 M in HNO3.
([HCOO-] = 9.0 x 10-5 M; pH = 1.00)
Sample Exercise 17.3 (p. 725)
What is the pH of a buffer that is 0.12 M in lactic acid (HC3H5O3) and 0.10 M in sodium lactate (Na C3H5O3)? For lactic acid, Ka = 1.4 x 10-4.
(3.77)
Practice Exercise 17.3
Calculate the pH of a buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. (Refer to Appendix D)
(4.42)
Sample Exercise 17.4 (p. 726)
How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH is 9.00? (Assume that the addition of NH4Cl does not change the volume of the solution.)
(0.36 mol)
Practice Exercise 17.4
Calculate the concentration of sodium benzoate that must be present in a 0.20 M solution of benzoic acid (HC7H5O2) to produce a pH of 4.00.
(0.13 M)
Sample Exercise 17.5 (p. 728)
A buffer is made by adding 0.300 mol of HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74 (Sample Exercise 17.1).
a) Calculate the pH of this solution after 0.020 mol of NaOH is added (4.80)
b) For comparison, calculate the pH that would result if 0.020 mol of NaOH were added to 1.00 L of pure
water. (neglect any volume changes). (12.30)
Practice Exercise 17.5
Determine
a) the pH of the original buffer described in Sample Exercise 17.5 after the addition of 0.020 mol HCl, and (4.68)
b) the pH of the solution that would result from the addition of 0.020 mol HCl to 1.00 L of pure water. (1.70)
Seven Steps to solving titration problems:
Analyze the problem, then:
1. Determine # moles of each reactant.
2. Determine # moles of all species after reaction.
3. Calculate new volume after reaction.
4. Determine molarities of all species (combine Steps 2 & 3).
5. Equilibrium calculation – RICE table. *
6. Equili brium calculation, substituting results from Step 5 into Ka. *
7. [H+] à pH. May mean [OH-] à pOH à 14.00 – pOH = pH.
*not required for strong acid-strong base titrations
Sample Exercise 17.6 (p. 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution:
a) 49.0 mL ( 3.00)
b) 51.0 mL (11.00)
Practice Exercise 17.6
Calculate the pH when the following quantities of 0.10 M HNO3 have been added to 25.0 mL of 0.10 M KOH solution:
a) 24.9 mL (10.30)
b) 25.1 mL ( 3.70)
Sample Exercise 17.7 (p. 735)
Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH solution is added to 50.0 mL of 0.100 M HC2H3O2. (Ka = 1.8 x 10-5)
(2.0 x 10-6 M)
Practice Exercise 17.7
a) Calculate the pH in the solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic acid (HC7H5O2, Ka = 6.3 x 10-5). (4.20)
b) Calculate the pH in the solution formed by adding 10.0 mL of 0.100 M HCl to 20.0 mL of 0.100 M NH3. (9.26)
Sample Exercise 17.8 (p. 735)
Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M HC2H3O2 with 0.100 M NaOH.
(8.72)
Practice Exercise 17.8
Calculate the pH at the equivalence point when
a) 40.0 mL of 0.025 M benzoic acid (HC7H5O2, Ka = 6.3 x 10-5) is titrated with 0.050 M NaOH (8.21)
b) 40.0 mL of 0.100 M NH3 is titrated with 0.100 HCl (5.28)
Sample Exercise 17.9 (p. 738)
Write the expression for the solubility-product constant for CaF2, and look up the corresponding Ksp value in Appendix D.
Practice Exercise 17.9
Give the solubility-product constant expressions and the values of the solubility-product constants (from Appendix D) for the following compounds:
a) barium carbonate
b) silver sulfate
Sample Exercise 17.10 (p. 739)
Solid silver chromate is added to pure water at 25oC. Some of the solid remains undissolved at the bottom of the flask. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag2CrO4(s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is
1.3 x 10-4 M. Assuming that Ag2CrO4 dissociates completely in water and that there are no other important equilibria involving the Ag+ or CrO42- ions in the solution, calculate Ksp for this compound.
(1.1 x 10-12)
Practice Exercise 17.10
A saturated solution of Mg(OH)2 in contact with undissolved solid is prepared at 25oC. The pH of the solution is found to be 10.17. Assuming that Mg(OH)2 dissociates completely in water and that there are no other simultaneous equilibria involving the Mg2+ or OH- ions in the solution, calculate Ksp for this compound.
(1.6 x 10-12)
Sample Exercise 17.11 (p. 740)
The Ksp for CaF2 is 3.9 x 10-11 at 25oC. Assuming that CaF2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF2 in grams per liter.
(1.6 x 10-2 g CaF2/L soln)
Practice Exercise 17.11
The Ksp for LaF3 is 2 x 10-19. What is the solubility of LaF3 in water in moles per liter?
(9 x 10-6 mol/L)
Sample Exercise 17.12 (p. 742)
Calculate the molar solubility of CaF2 at 25oC in a solution that is
a) 0.010 M Ca(NO3)2 (3.1 x 10-5 mol CaF2/L 0.010 M Ca(NO3)2)
b) 0.010 M in NaF (3.9 x 10-7 mol CaF2/L 0.010 M NaF)
Practice Exercise 17.12
The value for Ksp for manganese (II) hydroxide, Mn(OH)2, is 1.6 x 10-13. Calculate the molar solubility of Mn(OH)2 in a solution that contains 0.020 M NaOH.
(4.0 x 10-10 M)
Sample Exercise 17.13 (p. 745)
Which of the following substances will be more soluble in acidic solution than in basic solution:
a) Ni(OH)2(s)
b) CaCO3(s)
c) BaF2(s)
d) AgCl(s)
(a-c)
Practice Exercise 17.13
Write the net ionic equation for the reaction of the following copper (II) compounds with acid:
a) CuS
b) Cu(N3)2
Sample Exercise 17.14 (p. 748)
Calculate the concentration of Ag+ present in solution at equilibrium when concentrated ammonia is added to a 0.010 M solution of AgNO3 to give an equilibrium concentration of [NH3] = 0.20 M. Neglect the small volume change that occurs when NH3 is added.
([Ag+] = 1.5 x 10-8 M)
Practice Exercise 17.14
Calculate [Cr3+] in equilibrium with Cr(OH)4- when 0.010 mol of Cr(NO3)3 is dissolved in a liter of solution buffered at pH 10.0.
([Cr3+] = 1 x 10-16 M)
Sample Exercise 17.15 (p.751)
Will a precipitate form when 0.10 L of 8.0 x 10-3 M Pb(NO3)2 is added to 0.40 L of 5.0 x 10-3 M Na2SO4?
(yes)
Practice Exercise 17.15
Will a precipitate form when 0.050 L of 2.0 x 10-2 M NaF is mixed with 0.010 L of 1.0 x 10-2 M Ca(NO3)2?
(yes)
Sample Exercise 17.16 (p. 751)
A solution contains 1.0 x 10-2 M Ag+ and 2.0 x 10-2 M Pb2+. When Cl- is added to the solution, both AgCl (Ksp = 1.8 x 10-10) and PbCl2 (Ksp = 1.7 x 10-5) precipitate from the solution.
What concentration of Cl- is necessary to begin the precipitation of each salt?
Which salt precipitates first?
(> 2.9 x 10-2 M for PbCl2; > 1.8 x 10-8 M for AgCl, precipitates first)
Practice Exercise 17.16
A solution consists of 0.050 M Mg2+ and 0.020 M Cu2+.
Which ion will precipitate first as OH- is added to the solution?
What concentration of OH- is necessary to begin the precipitation of each cation?
(Ksp = 1.8 x 10-11 for Mg(OH)2 and Ksp = 4.8 x 10-20 for Cu(OH)2)
(Cu(OH)2 precipitates first, when [OH-] > 1.5 x 10-9 M; Mg(OH)2 precipitates when [OH-] > 1.9 x 10-5 M)
Sample Integrative Exercise 17 (p. 755)
A sample of 1.25 L of HCl gas at 21oC and 0.950 atm is bubbled through 0.500 L of 0.150 M NH3 solution. Calculate the pH of the resulting solution, assuming that all of the HCl dissolves and that the volume of the solution remains 0.500 L.
(pH = 8.97)
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